Introduction

In this blog post, we will explore how to solve the LeetCode problem "1742" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are working in a ball factory where you have n balls numbered from lowLimit up to highLimit inclusive (i.e., n == highLimit - lowLimit + 1), and an infinite number of boxes numbered from 1 to infinity. Your job at this factory is to put each ball in the box with a number equal to the sum of digits of the ball's number. For example, the ball number 321 will be put in the box number 3 + 2 + 1 = 6 and the ball number 10 will be put in the box number 1 + 0 = 1. Given two integers lowLimit and highLimit, return the number of balls in the box with the most balls. Example 1: Input: lowLimit = 1, highLimit = 10 Output: 2 Explanation: Box Number: 1 2 3 4 5 6 7 8 9 10 11 ... Ball Count: 2 1 1 1 1 1 1 1 1 0 0 ... Box 1 has the most number of balls with 2 balls. Example 2: Input: lowLimit = 5, highLimit = 15 Output: 2 Explanation: Box Number: 1 2 3 4 5 6 7 8 9 10 11 ... Ball Count: 1 1 1 1 2 2 1 1 1 0 0 ... Boxes 5 and 6 have the most number of balls with 2 balls in each. Example 3: Input: lowLimit = 19, highLimit = 28 Output: 2 Explanation: Box Number: 1 2 3 4 5 6 7 8 9 10 11 12 ... Ball Count: 0 1 1 1 1 1 1 1 1 2 0 0 ... Box 10 has the most number of balls with 2 balls. Constraints: 1 <= lowLimit <= highLimit <= 105

Explanation

Here's the breakdown of the solution:

• Calculate Digit Sum: Efficiently compute the sum of digits for each ball number within the given range.
• Count Box Frequencies: Use a dictionary (or array) to store the frequency of each digit sum (box number).

• Find Maximum Frequency: Iterate through the frequency counts to determine the maximum frequency, which represents the number of balls in the box with the most balls.

• Runtime Complexity: O(n), where n is the range (highLimit - lowLimit + 1).

• Storage Complexity: O(1), as the maximum possible digit sum is bounded (9 * number of digits). The number of digits is at most 6, resulting in a constant storage requirement.

Code

    def count_balls(lowLimit: int, highLimit: int) -> int:
    """
    Calculates the number of balls in the box with the most balls.

    Args:
        lowLimit: The lower limit of the ball numbers.
        highLimit: The upper limit of the ball numbers.

    Returns:
        The number of balls in the box with the most balls.
    """

    box_counts = {}
    max_count = 0

    for ball_number in range(lowLimit, highLimit + 1):
        digit_sum = 0
        temp = ball_number
        while temp > 0:
            digit_sum += temp % 10
            temp //= 10

        if digit_sum not in box_counts:
            box_counts[digit_sum] = 0
        box_counts[digit_sum] += 1

        max_count = max(max_count, box_counts[digit_sum])

    return max_count